CODEWITHSUNDEEP
javascript
kate
kate

Reputation: 13

backgroundColor as switch function?

This code shows no errors but is not working, can anyone see why?

function setStyles(){
    var ya
    styles=document.getElementById("back");
    switch (ya){
        case 1:styles.style.backgroundColor='#FF0000';
            break;
        case 2:styles.style.backgroundColor='#FF6600';
            break;
        case 3:styles.style.backgroundColor='#FFFF66';
            break;
        case 4:styles.style.backgroundColor='#669900';
            break;
        case 5:styles.style.backgroundColor='#000066';
            break;
        case 6:styles.style.backgroundColor='#660066';
            break;
    }
}

window.onload = setStyles;

Upvotes: 0

Views: 48

Answers (2)

MrCode
MrCode

Reputation: 64536

None of the case statements will be matched because ya is undefined. Try defining it (in other words give it a value):

var ya = 1;

You could also define a default case which will be executed if none of the other case statements match:

...
...
case 6:
    styles.style.backgroundColor='#660066';
    break;
default:
    styles.style.backgroundColor='#000000';
...

Edit: to get a random background color:

var ya = Math.floor(Math.random() * 6) + 1;

Upvotes: 0

Charles
Charles

Reputation: 11788

It works if you define ya:

function setStyles(ya){
  var ya;

  styles=document.getElementById("back");

  switch (ya) {
    case 1:
      styles.style.backgroundColor='#FF0000';
    break;

    case 2:
      styles.style.backgroundColor='#FF6600';
    break;

    case 3:
      styles.style.backgroundColor='#FFFF66';
    break;

    case 4:
      styles.style.backgroundColor='#669900';
    break;

    case 5:
      styles.style.backgroundColor='#000066';
    break;

    case 6:
      styles.style.backgroundColor='#660066';
    break;
  }
}

window.onload = function() {
  setStyles(1);
};

http://jsbin.com/agajen/1/edit

Upvotes: 1

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